Pku oj 1905 Expanding Rods（简单几何加二分）_p4058 [code+#1]木材-优快云博客

本文链接：https://blog.youkuaiyun.com/Sara_YF/article/details/51999502

本文介绍了一种计算受热膨胀的细长杆发生弯曲时中心位移的方法。通过数学模型将问题转化为几何问题，并采用二分查找算法求解精确位移。输入包括杆的初始长度、温度变化及材料膨胀系数。

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Expanding Rods

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 249(100 users)

Total Accepted: 121(93 users)

Rating:

Special Judge: No

Description

When a thin rod of length L is heated n degrees, it expands to a new length L'=(1+n*C)*L, where C is the coefficient of heat expansion.
When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment.

Your task is to compute the distance by which the center of the rod is displaced.

Input

The input contains multiple lines. Each line of input contains three non-negative numbers: the initial lenth of the rod in millimeters, the temperature change in degrees and the coefficient of heat expansion of the material. Input data guarantee that no rod expands by more than one half of its original length. The last line of input contains three negative numbers and it should not be processed.

Output

For each line of input, output one line with the displacement of the center of the rod in millimeters with 3 digits of precision.

Sample Input

1000 100 0.0001
15000 10 0.00006
10 0 0.001
-1 -1 -1

Sample Output

61.329
225.020

0.000

几何方式求L’，然后再与用题中公式求得进行比较，以此来二分答案

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;

const double eqs = 1e-8;
double l,n,s,c;
double h;
double r;

int main()
{
    while(~scanf("%lf%lf%lf",&l,&n,&c))
    {
        if(l < 0 && n < 0 && c < 0) break;
        s = (1 + n * c) * l;
        double low = 0,high = 0.5 * l,mid;
        while(high - low > eqs)
        {
            mid = (high + low) / 2;
            r = (4*mid*mid + l*l) / (8*mid);
            if(2*r*asin(l/(2*r)) < s)
                low = mid;
            else
                high = mid;
        }
        printf("%.3lf\n",mid);
    }
}